Adjoint (operator theory)

In mathematics, the adjoint of an operator is a generalization of the notion of the Hermitian conjugate of a complex matrix to linear operators on complex Hilbert spaces. In this article the adjoint of a linear operator M will be indicated by M^∗, as is common in mathematics. In physics the notation M^† is more usual.

Main idea

Consider a complex n×n matrix M. Apart from being an array of complex numbers, M can also be viewed as a linear map or operator from ℂⁿ to itself. In order to generalize the idea of the Hermitian conjugate of a complex matrix to linear operators on more general complex Hilbert spaces, it is necessary to be able to characterize the Hermitian conjugate as an operator. The crucial observation here is the following: for any complex matrix M, its Hermitian tranpose, denoted by M^∗, is the unique linear operator on ℂⁿ satisfying:

$\langle Mx,y\rangle =\langle x,M^{*}y\rangle \quad \forall x,y\in \mathbb {C} ^{n}.$

This suggests that the "Hermitian conjugate" or, as it is more commonly known in mathematics, the adjoint of a linear operator T on an arbitrary complex Hilbert space H, with inner product ⟨ ⋅, ⋅ ⟩_H, could be defined generally as an operator T^∗ on H satisfying the so-called "turn-over rule":

$\langle Tx,y\rangle _{H}=\langle x,T^{*}y\rangle _{H}\qquad \forall x,y\in H.\qquad \qquad \qquad \qquad (1)$

It turns out that this idea is almost correct. It is correct and a unique T^∗ exists, if T is a bounded operator on H, but additional care has to be taken on infinite dimensional Hilbert spaces since operators on such spaces can be unbounded and there may not exist an operator T^∗ satisfying (1).

Existence of the adjoint

Suppose that T is a densely defined operator on H with domain D(T). Consider the vector space

$K(T)=\{\;v\in H\;\mid \;{\underset {u\in D(T)}{\sup }}|\langle Tu,v\rangle _{H}|<\infty \;\},$

that is, the space consists of all vectors v of which the supremum of the absolute value of ⟨Tu, v ⟩_H is finite. Since T has a dense domain in H and $f_{v}(u)\equiv \langle Tu,v\rangle _{H}$ is a continuous linear functional on D(T) for any v ∈ K(T), f_v can be extended to a unique continuous linear functional ${\tilde {f}}_{v}$ on H. By the Riesz representation theorem there is a unique element v^∗ ∈ H such that

${\tilde {f}}_{v}(u)=\langle u,v^{*}\rangle _{H}\quad \forall u\in H.$

A linear operator T^∗ with domain D(T^∗) = K(T) may now be defined as the map

$T^{*}v=v^{*}\quad \forall v\in D(T).$

By construction, the operator T^∗ satisfies:

$\langle Tx,y\rangle _{H}=\langle x,T^{*}y\rangle _{H}\qquad \forall x\in D(T),\quad \forall y\in D(T^{*}).\qquad \qquad \qquad \qquad (2)$

When T is a bounded operator (hence D(T) = H) then it can be shown, again using the Riesz representation theorem, that T^∗ is the unique bounded linear operator satisfying equation (2).

Formal definition of the adjoint of an operator

Let T be an operator on a Hilbert space H with dense domain D(T). Then the adjoint T^∗ of T is an operator with domain

$D(T^{*})=\{\;v\in H\mid {\underset {u\in D(T)}{\sup }}|\langle Tu,v\rangle _{H}|<\infty \;\}$

defined as the map

$T^{*}v=v^{*}\quad \forall v\in D(T^{*}),$

where for each v in D(T^∗), v^∗ is the unique element of H such that

$\langle u,v^{*}\rangle =\langle Tu,v\rangle _{H}\quad \forall u\in D(T).$

Additionally, if T is a bounded operator then T^∗ is the unique bounded operator satisfying

$\langle Tx,y\rangle _{H}=\langle x,T^{*}y\rangle _{H}\quad \forall x,y\in H.$

Property

Consider two linear operators S and T on H with overlapping domains. For convenience we assume D(T) = D(S) and D(T^∗) = D(S^∗). Then

$\langle \;aTS\,u,\;v\;\rangle _{H}=\langle \;u,\;(aTS)^{*}\,v\;\rangle _{H}=\langle \;u,\;{\overline {a}}S^{*}T^{*}\,v\;\rangle _{H},\quad a\in \mathbb {C} .$

Proof

The fact that the complex conjugate of the complex number a appears is due to the property of the inner product on complex Hilbert space. The fact that the multiplication order of the operators reverts under the turnover rule follows thus

$\langle TS(u),v\rangle _{H}=\langle Tu',v\rangle _{H}=\langle u',T^{*}(v)\rangle _{H}=\langle u',v'\rangle _{H}=\langle S(u),v'\rangle _{H}=\langle u,S^{*}v'\rangle _{H}=\langle u,S^{*}\,T^{*}(v)\rangle _{H},$

with

$u'\equiv S(u),\quad v'\equiv T^{*}(v).$